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I have some questions about some arguments used in the discussion about consequnces of Bott periodity in in A Concise Course in Algebraic Topology by P. May at page 207. Her the excerpt:

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Following context: Denote $U:= colim_n U(n)$ the infinite unitary group induced via canonical inclusions $U(n) \subset U(n+1)$.

FIRST QUESTION: Why is the loop space $\Omega BU := Hom(S^1, BU)$ $H$- equivalent to $U$?

We know that $BU = E/U$ where $E$ is the unique simply connected universal principal $U$-space and after applying $\pi_0(-)$ functor we obtain $\pi_0(\Omega BU) = \pi_1(E/U)= U$ by covering theory since $E$ is universal cover.

But does this already imply that $\Omega BU $ homotopy equivalent to $U$? $\pi_0$ only counts path components.

SECOND QUESTION: We know that $SU$ - the infinite special unitary group - is the universal cover of $U$ (via colimit argument). Futhermore $\pi_1(U)=\mathbb{Z}$.

Why does it imply that $$\Omega U \cong (\Omega SU) \times \mathbb{Z}$$ as $H$-spaces (remark: $X$ is a $H$-space if there exist a "multiplication" $X \times X \to X$.

user267839
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1 Answers1

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Consider the map of fibrations from the universal principal $U$-bundle over $BU$ to the path-loop fibration on $BU$:

$\require{AMScd}$ \begin{CD} U @>>> EU @>>> BU \\ @VVV @VVV @| \\ \Omega BU @>>> PBU @>>> BU \end{CD}

One can write down a map $EU \to PBU$; thus there is a map $U \to \Omega BU$. Looking at the long exact sequence in homotopy and using the five-lemma, one deduces that the induced map $U \to \Omega BU$ is a weak homotopy equivalence because $EU \simeq PBU \simeq *$. (Since both source and target can be equipped with a CW structure, this can be promoted to a homotopy equivalence.)

In general, this shows $\Omega BG \simeq G$ for a topological group $G$. See this answer for more details.

For your second question, there is a short exact sequence of topological groups $$1 \to SU \to U \xrightarrow{\det} S^1 \to 1$$ which is split, so $U \cong SU \rtimes S^1$ as topological groups. If we ignore the group structure and only care about the topology, then we have a homeomorphism $U \cong SU \times S^1$. Taking loops on both sides, we get $$\Omega U \cong \Omega SU \times \Omega S^1 \simeq \Omega SU \times \mathbb{Z}$$ as loop spaces, hence as $H$-spaces.

JHF
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  • Thank you for answer. One point iritates me: In the last line you claim (neglecting the left factor $\Omega SU$) that $\Omega S^1 \simeq \mathbb{Z}$. Why does it hold? I think that it only holds after passing to homtopy groups. More precisely we know that the loop space is defined as $\Omega S^1 := Hom(S^1, S^1)$. On the other hand we know that for *homotopy classes* holds indeed $[S^1,S^1]=\mathbb{Z}$ where the conection is the path component functor $\pi_0(\Omega S^1 )= [S^1,S^1]$. But I'm not sure why should hold $\Omega S^1 \simeq \mathbb{Z}$ *before* passing to homotopy classes? – user267839 Mar 23 '19 at 00:27
  • Another remark: Do I understand the argument with the $H$-story correctly that already $U \cong SU \times S^1$ suffice since $\Omega(-)$ endows in a functorial way the spaces with $H$-space structure? And therefore $H$-space property is here equivalent to functoriality of $\Omega(-)$? – user267839 Mar 23 '19 at 00:27
  • $\Omega S^1$ is homotopy discrete (it has no higher homotopy groups, because $S^1$ has no homotopy groups above degree $1$). For your second question: if $X \cong Y$, then $\Omega X \cong \Omega Y$ as loop spaces. You might be concerned about the original multiplication on $X$ and $Y$ if they were groups to begin with, but an Eckmann-Hilton argument would show that the operations coincide. Finally, note that a loop space equivalence is stronger than a $H$-space equivalence. – JHF Mar 25 '19 at 16:07
  • hmmm so for $\Omega S^1 \simeq \mathbb{Z}$ you also implicitely use Whitehead, don't you? https://en.wikipedia.org/wiki/Whitehead_theorem

    Since your homotopy discreteness argument provides that $\Omega S^1$ and $\coprod_{z \in \mathbb{Z}} {*} \cong \mathbb{Z}$ have the same homotopy groups. Or did I misunderstood your point?

     – user267839 Mar 25 '19 at 16:48
  • Sure that works. Or you could argue $\Omega S^1 \simeq \Omega K(\mathbb{Z}, 1) \simeq K(\mathbb{Z}, 0) \simeq \mathbb{Z}$, which is basically the same argument. – JHF Mar 25 '19 at 17:12